Answer in Statistics and Probability for Biostat101 #250482
The average running time for Broadway shows is 2 hours and 12 minutes. A producer in another city claims that the length of time of productions in his city is the same. He samples 8 shows and finds the time to be 2 hours and 5 minutes with a standard deviation of 11 minutes. Using , is the producer correct?
The following null and alternative hypotheses need to be tested:
“H_0:mu=132 min”
“H_1:munot=132 min”
This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.
Based on the information provided, the significance level is “alpha = 0.05, df=n-1”
“=8-1=7” degrees of freedom, and the critical value for a two-tailed test is“t_c =2.364619.”
The rejection region for this two-tailed test is “R = {t: |t| > 2.364619}.”
The t-statistic is computed as follows:
“t=dfrac{bar{x}-mu}{s/sqrt{n}}=dfrac{125-132}{11/sqrt{8}}=-1.7999”
Since it is observed that “|t| = 1.7999<2.364619=t_c,” it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value for two-tailed, “alpha=0.05, df=7,” “t=-1.7999” is “p= 0.114901,” and since “p= 0.114901>0.05=alpha,” it is concluded that the null hypothesis is not rejected.
Therefore, there is not enough evidence to claim that the population mean “mu” is different than 132, at the “alpha = 0.05” significance level.
Therefore, there is enough evidence to claim that the producer is correct, at the “alpha = 0.05” significance level.
